有條MI唔識做...求助 [復制鏈接]

引用第13樓樂仔於2007-06-03 19:19發表的“”:
點解我會計到咁既?
(k+2)^2+(k+3)^2+(k+4)^2+...+(2k)^2+(2k+2)^2
中間少左個(2k+1)^2既????
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你要明左條式係點行先
佢係由1開始, 1下+1, 1,2,3,4,5,...,n-2, n-1, n

條式最後係 (2n)^2
當put左k+1入去
即最後係 [2(k+1)]^2
i.e, (2k+2)^2
由於條式係+1,+1咁上
咁(2k+2)^2之前係有(2k+1)^2, (2k)^2, (2k-1)^2,....

引用第14樓樂仔於2007-06-03 19:20發表的“”:
= {[k(2k+1)(7k+1)]/6 - (k+1)^2} + (2K+1)^2 + (2k+2)
點解要減(k+1)^2? [表情]
即係將上一步冇寫到既(k+1)^2搬返落黎咁解?

(n+1)^2+(n+2)^2+(9n+3)^2+...+(2n)^2 = [n(2n+1)(7n+1)]/6

因為put左n=k+1, 就冇左第一舊

(n+1)^2 + (n+2)^2+(9n+3)^2+...+(2n)^2 = [n(2n+1)(7n+1)]/6

(n+2)^2+(9n+3)^2+...+(2n)^2 = [n(2n+1)(7n+1)]/6 - (n+1)^2

(n+2)^2+(9n+3)^2+...+(2n)^2 + [(2n+1)^2 + (2n+2)^2] = [n(2n+1)(7n+1)]/6 - (n+1)^2 + [(2n+1)^2 + (2n+2)^2]

引用第16樓ballballfull於2007-06-03 19:30發表的“”:
(n+1)^2+(n+2)^2+(9n+3)^2+...+(2n)^2 = [n(2n+1)(7n+1)]/6
因為put左n=k+1, 就冇左第一舊
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明白

哦...明哂啦!